1. ⇒ (MHT CET 2023 13th May Morning Shift )

A circular coil of radius ' $r$ ' and number of turns ' $n$ ' carries a current ' $I$ '. The magnetic fields at a small distance ' $h$ ' along the axis of the coil $(B_{a})$ and at the centre of the coil $(B_{c})$ are measured. The relation between $B_{c}$ and $B_{a}$ is

A. $B_{c} = B_{a} (1 + \frac{h^{2}}{r^{2}})$

B. $B_{c} = B_{a} {(1 + \frac{h^{2}}{r^{2}})}^{\frac{1}{2}}$

C. $B_{c} = B_{a} {(1 + \frac{h^{2}}{r^{2}})}^{\frac{3}{2}}$

D. $B_{c} = B_{a} {(1 + \frac{h^{2}}{r^{2}})}^{- \frac{3}{2}}$

Correct Option is (C)

Magnetic field along the axis of the coil is:

$B_{a} = \frac{μ_{0}}{4 π} [\frac{2 π n {Ir}^{2}}{{(r^{2} + h^{2})}^{\frac{3}{2}}}]$ .... (i)

Magnetic field at the centre of the coil is:

$\begin{aligned} B_{C} & = \frac{μ_{0}}{4 π} (\frac{2 π n I}{r}) .... (ii) \\ \frac{B_{C}}{B_{a}} & = \frac{{(r^{2} + h^{2})}^{\frac{3}{2}}}{r^{3}} \\ ∴ B_{C} & = B_{a} {[1 + \frac{h^{2}}{r^{2}}]}^{\frac{3}{2}} \end{aligned}$

2. ⇒ (MHT CET 2023 13th May Morning Shift )

Two concentric circular coils A and B have radii $20 cm$ and $10 cm$ respectively lie in the same plane. The current in coil A is $0.5 A$ in anticlockwise direction. The current in coil B so that net field at the common centre is zero, is

A. $0.5 A$ in anticlockwise direction

B. $0.25 A$ in anticlockwise direction.

C. $0.25 A$ in clockwise direction.

D. $0.125 A$ in clockwise direction.

Correct Option is (D)

Magnetic field at the centre of a circular loop

$B = \frac{μ_{0} NI}{2 R}$

Net Magnetic field $B_{net} = 0$ (given)

$\begin{aligned} \Rightarrow & B_{net} = \frac{μ_{0} N_{1} \times 0.5}{2 \times (0 \cdot 2)} - \frac{μ_{0} N_{1} \times x}{2 \times (0 \cdot 1)} \\ \Rightarrow & \frac{μ_{0} N_{1} x}{0.2} = \frac{μ_{0} N_{1} 0.5}{0.4} \\ ∴ x & = \frac{0.5 \times 0.2}{0.4} \\ = 0.25 A in clockwise direction \end{aligned}$

3. ⇒ (MHT CET 2023 12th May Evening Shift )

Two concentric circular coils of 10 turns each are situated in the same plane. Their radii are $20 cm$ and $40 cm$ and they carry respectively $0.2 A$ and $0.3 A$ current in opposite direction. The magnetic field at the centre is ( $μ_{0} = 4 π \times 10^{- 7}$ SI units)

A. $4 π \times 10^{- 7} T$

B. $5 π \times 10^{- 7} T$

C. $2 π \times 10^{- 5} T$

D. $7 π \times 10^{- 6} T$

Correct Option is (B)

$\begin{aligned} B_{net} & = \frac{μ_{0} n_{1} I_{1}}{2 r_{1}} - \frac{μ_{0} n_{2} I_{2}}{2 r_{2}} \\ = \frac{10 μ_{0}}{2} [\frac{0.2}{0.2} - \frac{0.3}{0.4}] \\ = 5 \times 4 π \times 10^{- 7} \times 0.25 \\ = 5 π \times 10^{- 7} T \end{aligned}$

4. ⇒ (MHT CET 2023 12th May Evening Shift )

An electron makes a full rotation in a circle of radius $0.8 m$ in one second. The magnetic field at the centre of the circle is $(μ_{0} = 4 π \times 10^{- 7}$ SI units)

A. $4 π \times 10^{- 26} T$

B. $2 π \times 10^{- 26} T$

C. $4 π \times 10^{- 19} T$

D. $2 π \times 10^{- 19} T$

Correct Option is (A)

$\begin{aligned} ω = \frac{2 π}{T} \\ I = \frac{q ω}{2 π} = \frac{1.6 \times 10^{- 19} \times 2 π}{2 π} \dots . (∵ I = \frac{q}{t}) \\ I = 1.6 \times 10^{- 19} A \end{aligned}$

$∴$ The magnetic field at the centre of the circle is:

$\begin{aligned} B = \frac{μ_{0} I}{2 r} = \frac{4 π \times 10^{- 7} \times 1.6 \times 10^{- 19}}{2 \times 0.8} \\ B = 4 π \times 10^{- 26} T \end{aligned}$

5. ⇒ (MHT CET 2023 12th May Morning Shift )

Two similar coils each of radius $R$ are lying concentrically with their planes at right angles to each other. The current flowing in them are I and 2I. The resultant magnetic field of induction at the centre will be $(μ_{0} =$ Permeability of vacuum)

MHT CET 2023 12th May Morning Shift Physics - Moving Charges and Magnetism Question 4 English

A. $\frac{μ_{0} I}{2 R}$

B. $\frac{μ_{0} I}{R}$

C. $\frac{3 μ_{0} I}{2 R}$

D. $\frac{\sqrt{5} μ_{0} I}{2 R}$

Correct Option is (D)

$B_{1} = \frac{μ_{0} I}{2 R}, B_{2} = \frac{μ_{0} (2 I)}{2 R}$

Resultant magnetic field, $B = \sqrt{B_{1} + B_{2}}$

$\begin{array}{ll} ∴ & B = \sqrt{{(\frac{μ_{0} I}{2 R})}^{2} + {(\frac{μ_{0} (2 I)}{2 R})}^{2}} \\ ∴ & B = \frac{μ_{0} I}{2 R} \sqrt{1^{2} + 2^{2}} \\ ∴ & B = \frac{\sqrt{5} μ_{0} I}{2 R} \end{array}$

⇒Magnetic Field due to Electric Current